Problem: You have found the following ages (in years) of all $4$ bears at your local zoo: $ 5,\enspace 4,\enspace 6,\enspace 39$ What is the average age of the bears at your zoo? What is the standard deviation? Round your answers to the nearest tenth. Average age: $ $
Solution: Because we have data for all $4$ bears at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$. To find the population mean, add up the values of all $4$ ages and divide by $4$. $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{5 + 4 + 6 + 39}{{4}} = {13.5\text{ years old}} $ Find the squared deviations from the mean for each bear. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $5$ years $-8.5$ years $72.25$ years $^2$ $4$ years $-9.5$ years $90.25$ years $^2$ $6$ years $-7.5$ years $56.25$ years $^2$ $39$ years $25.5$ years $650.25$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean, we can find the variance $({\sigma^2})$, without introducing any bias, by simply averaging the squared deviations from the mean: $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{72.25} + {90.25} + {56.25} + {650.25}} {{4}} $ $ {\sigma^2} = \dfrac{{869}}{{4}} = {217.25\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$. ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{217.25\text{ years}^2}} = {14.7\text{ years}} $ The average bear at the zoo is $13.5$ years old. There is a standard deviation of $14.7$ years.